Chapter 15, Problem 108AP

Consider the equilibrium reaction described in Problem 15.30. A quantity of 2.50 g of ${\text{PCl}}_{\text{5}}$. is placed in an evacuated 0.500-L flask and heated to $\text{250°C}$. (a) Calculate the pressure of ${\text{PCl}}_{\text{5}}$, assuming it does not dissociate, (b) Calculate the partial pressure of ${\text{PCl}}_{\text{5}}$ at equilibrium, (c) What is the total pressure at equilibrium? (d) What is the degree of dissociation of ${\text{PCl}}_{\text{5}}$ ? (The degree of dissociation is given by the fraction of ${\text{PCl}}_{\text{5}}$ that has undergone dissociation.)

Interpretation Introduction

Interpretation:

The pressure of $PC{l}_5$, if it does not dissociate, the partial pressure of $PC{l}_5$ at equilibrium, the total pressure at equilibrium, and the degree of dissociation of $PC{l}_5$ in the reaction are to be calculated.

Concept introduction:

According to Dalton’s law, the total pressure of a mixture of two or more nonreactive gases is equal to the individual partial pressure of the gases present in the mixture. The pressure exerted by a gas in the mixture of gases is called its partial pressure.

The expression given by the ideal gas equation is as follows:

$PV=nRT$

$\text{Number of mole (}n)=\frac{\text{Mass of compound}}{\text{Molar mass of compound}}$

Answer to Problem 108AP

Solution:

$1.03\text{atm}$

$0.39\text{atm}$

$1.67\text{atm}$

$0.620$

Explanation of Solution

a)Pressure of ${\text{PCl}}_5$, assuming it does not dissociate

The equilibrium reaction is given as:

${\text{PCl}}_5\left(g\right)\rightleftarrows {\text{PCl}}_3\left(g\right)+{\text{Cl}}_2$

The mass of ${\text{PCl}}_5$ in the reaction is $2.50\text{g}$.

The volume of the flask is $0.500\text{L}$.

The temperature (T) of the reaction is $250°\text{C}$.

Temperature $=273+{250}^0\text{C=523}\text{K}$

The molar mass of $PC{l}_5$ is $208.2\text{g}/\text{mol}$.

Number of moles $PC{l}_5$ $n=2.50\text{g}\times \frac{\text{1}\text{mol}}{\text{208}\text{.2}\text{g}}$

The pressure of $PC{l}_5$ is calculated by the expression as follows:

$\begin{array}{c}PV=nRT\\ P=\frac{nRT}{V}\end{array}$

Here, $P$ is the pressure, $n$ is the number of moles, $R$ is the gas constant, $T$ is the temperature, and $V$ is the volume.

Substitute the values of $n$, $R$, $T$, and $V$ in the equation given above,

$\begin{array}{c}P=\frac{\left(2.50\cancel{\text{g}}\times \frac{\text{1}\cancel{\text{mol}}}{\text{208}\text{.2}\cancel{\text{g}}}\right)\left(0.0821\times \frac{\cancel{\text{L}}\text{.atm}}{\cancel{\text{mol}}\text{.}\cancel{\text{K}}}\right)\left(523\cancel{\text{K}}\right)}{0.500\cancel{\text{L}}}\\ =\frac{\left(0.0120\right)\left(0.0821\right)\left(523\right)}{0.500}\text{atm}\\ =\frac{0.5156}{0.500}\text{atm}\\ =1.03\text{atm}\end{array}$

Hence, the pressure of $PC{l}_5$ is $1.03\text{atm}$.

b) Partial pressure of $PC{l}_5$ at equilibrium

The chemical reaction is given as follows:

${\text{PCl}}_5\left(g\right)\rightleftarrows {\text{PCl}}_3\left(g\right)+{\text{Cl}}_2$

The Initial change equilibrium table for the reaction is given as follows:

$\begin{array}{cccc}& {\text{PCl}}_{\text{5}}& {\text{PCl}}_{\text{3}}& {\text{Cl}}_{\text{2}}\\ \text{Initial}\left(\text{atm}\right)& 1.03& 0& 0\\ \text{Change}\left(\text{atm}\right)& -x& +x& +x\\ \text{Equilibrium}\left(\text{atm}\right)& 1.03-x& x& x\end{array}$

The partial pressure of $PC{l}_5$ at equilibrium is calculated by the expression as follows:

$K_p=\frac{P_{{\text{PCl}}_3}{P}_{{\text{Cl}}_2}}{P_{{\text{PCl}}_5}}$

Substitute the values of $P_{{\text{PCl}}_3},P_{{\text{PCl}}_5}$, and $P_{{\text{Cl}}_2}$ in the equation given above,

$\begin{array}{c}{K}_p=\frac{x^2}{\text{1}\text{.03-}x}\\ \text{1}\text{.05}=\frac{x^2}{\text{1}\text{.03-}x}\\ x^2+\text{1}\text{.05}-\text{1}\text{.08}=\text{0}\\ x=\text{0}\text{.639}\text{atm}\end{array}$

The partial pressure of $PC{l}_5$ is calculated as follows:

$\begin{array}{c}{P}_{{\text{PCl}}_{\text{5}}}=1.03\text{atm}-0.639\text{atm}\\ =0.39\text{atm}\end{array}$

The partial pressure of ${\text{PCl}}_{\text{5}}$ is $0.39\text{atm}$.

c) Total pressure at equilibrium

The chemical reaction is given as follows:

${\text{PCl}}_5\left(g\right)\rightleftarrows {\text{PCl}}_3\left(g\right)+{\text{Cl}}_2$

The Initial change equilibrium table for the reaction is given as follows:

$\begin{array}{cccc}& {\text{PCl}}_{\text{5}}& {\text{PCl}}_{\text{3}}& {\text{Cl}}_{\text{2}}\\ \text{Initial}\left(\text{atm}\right)& 1.03& 0& 0\\ \text{Change}\left(\text{atm}\right)& -0.639& +0.639& +0.639\\ \text{Equilibrium}\left(\text{atm}\right)& 1.03-0.639& 0.639& 0.639\end{array}$

Total pressure of the reaction at equilibrium is calculated as follows:

$\begin{array}{c}{P}_T=\left(1.03-0.639\right)\text{atm}+0.639\text{atm}+0.639\text{atm}\\ =\left(1.03+0.639\right)\text{atm}\\ =1.67\text{atm}\end{array}$

Total pressure of $PC{l}_5$ is $1.67\text{atm}$.

d) Degree of dissociation of $PC{l}_5$

The degree of dissociation of $PC{l}_5$ is calculated by the expression as follows:

$\text{Degree}\text{of}\text{dissociation}\text{of}{\text{PCl}}_{\text{5}}=\frac{\text{total pressure}}{\text{partial pressure}}$

$\begin{array}{c}\text{Degree}\text{of}\text{dissociation}\text{of}{\text{PCl}}_{\text{5}}=\frac{0.639\text{atm}}{1.03\text{atm}}\\ =0.620\end{array}$

The degree of dissociation of $PC{l}_5$ is $0.620$.